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If Bluenose is guilty then no witness is lying unless he is fearful. There is a witness who is fearful. Therefore, Bluenose is not guilty.

$$B\to[(\forall x)(\,(Wx\land\lnot Fx)\to(\lnot Lx)\,)] \tag{1}$$

$$(\exists x)(Wx\land Fx) \tag{2}$$

$$\lnot B \tag{3}$$

Is this interpretation correct ?

I am asked to deduce the conclusion which is $\lnot B$ in this case.

** T.E stands for tautological equivalence

My try is:

$$3.\quad \lnot B \lor [(\forall x)(\,(Wx\land \lnot Fx)\to (\lnot Lx)\,)] \quad1 \; T.E\\4. \quad\lnot B \lor [\lnot (\exists x)\lnot(\,(Wx\land \lnot Fx)\to (\lnot Lx)\,)] \quad \; 3Q1$$

I am not able to deduce the conclusion, I think I misinterpret the sentence.

The following is my last try:

$$\tag{1} B\to (\nexists x)(Wx\land Lx)\oplus (\exists x)(Wx\land Fx) \qquad P$$ $$\tag{2} (\exists x)(Wx\land Fx) \qquad P$$ $$\tag{3} B\to (\exists x)(Wx\land Lx) \qquad 1,2$$

I am still struggling.

Bram28
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Ali
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    Above all, the syllogism is invalid – Hagen von Eitzen Jul 09 '18 at 05:57
  • The formalization of 1 is wrong. It must be : $B \to ¬∃x [Wx ∧ (¬Fx \to Lx)]$. – Mauro ALLEGRANZA Jul 09 '18 at 06:11
  • I think it is equivalent to say that every witness who is not fearful is not lying. Yours, there is no witness who is not fearful lying. – Ali Jul 09 '18 at 06:58
  • I your formalize correctly 1, the conclusion easily follows assuming $B$ and deriving a contradictiion. – Mauro ALLEGRANZA Jul 09 '18 at 12:36
  • I agree with @HagenvonEitzen . Consider the situation where Bluenose is guilty, there is exactly one witness, and the witness is fearful. (It doesn't matter whether the witness is lying or not.) Then the hypotheses (I mean the English hypotheses, never mind their formalization) are both satisfied, but the alleged conclusion is false. – Andreas Blass Jul 10 '18 at 00:39

2 Answers2

2

There is some English word play going on. I think the author means it as follows.

No witness is lying unless the witness is fearful.

To work out the "unless" operator: $$A\mbox{ unless }B \equiv \neg A\rightarrow B \qquad (\equiv A\lor B)$$ is how it should work. In this topic the unless operator is also tackled.

Accordingly, sentence 1. translates to $$B\rightarrow \neg\exists x\left (Wx\ \land\ (\neg Lx\rightarrow Fx)\right ) \equiv B\rightarrow \neg\exists x(Wx\ \land\ (Lx\lor Fx))$$

To deduce $\neg B$ it suffices to show that the implication $$[B\rightarrow \neg\exists x(Wx\ \land\ (Lx\lor Fx))]\land [\exists x (Wx\land Fx)]\rightarrow \neg B\tag{i} $$ is a tautology.

For the sake of completeness. Assume implication (i) is false. Then we have two conflicting conditions $$\neg\exists x(Wx\ \land\ (Lx\lor Fx))\quad\mbox{and}\quad \exists x(Wx\ \land\ Fx). $$ The left hand condition is equivalent to $\forall x(\neg Wx\ \lor\ (\neg Lx\ \land\ \neg Fx))$. The right hand condition provides for some $a$ that $Wa\ \land\ Fa$ is true ( i.e there exists a fearful witness). The left hand condition, however, states that all witnesses are truthful and fearless, in particular witness $a$. This is impossible.

AlvinL
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  • Actually, it is an exercise in a book, so I am not really sure if something is missed or not, but my target is to check whether my interpretation of 1 is correct or not. – Ali Jul 09 '18 at 07:54
  • I'm confident I've given you correct 1., but I made an error in trying to deduce $\neg B$. The exercise is ok, $\neg B$ can be deduced from these facts. – AlvinL Jul 09 '18 at 07:59
  • Still, I cannot – Ali Jul 10 '18 at 00:19
  • 1- I really appreciate what you have done. 2- How do you know if unless is used in the weak sense or strong sense ? – Ali Jul 16 '18 at 13:13
  • @Ali I might have fallen victim to English word play. David K's answer is more convincing. I don't understand what you mean by 2. – AlvinL Jul 16 '18 at 13:30
  • 1- "unless" can be used in two different ways as far as I know. $\$ 2- if I apply what have said that $A$ unless $B, A:$ no witness is lying $(\forall x)(Wx \to \lnot Lx), B:$ he is fearful (it is equivalent to saying that there is a witness who is fearful) $, (\exists x)(Wx \land Fx). \$ $B \to [(\exists x)(Wx \land Lx) \to (\exists x)(Wx \land Fx)] \tag{1}$ $(\exists x)(Wx \land Fx) \tag{2} $. which tells nothing about B. – Ali Jul 16 '18 at 13:41
  • I'd have to agree that the exercise is poorly worded. I am trying to make no extra assumptions about the English language itself. I wasn't sure about the grouping of statements, either. – AlvinL Jul 16 '18 at 13:50
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.. no witness is lying unless he is fearful.

The common English meaning of this sentence is satisfied if every witness tells the truth. It also is satisfied if some witnesses lie, but every lying witness is fearful.

In short, this clause says that every witness either tells the truth or is fearful:

$$ (\forall x)(Wx \to (\lnot Lx \lor Fx)).$$

We put this together with "If Bluenose is guilty" and we have

$$ B \to (\forall x)(Wx \to (\lnot Lx \lor Fx)). \tag1$$

Now, $(Wx \to (\lnot Lx \lor Fx)) \equiv (\lnot Wx \lor (\lnot Lx \lor Fx)).$ Therefore $(1)$ is equivalent to the following:

$$ B \to (\forall x)(\lnot Wx \lor \lnot Lx \lor Fx). \tag{$1'$}$$

Also, $((Wx\land\lnot Fx)\to(\lnot Lx)) \equiv (\lnot(Wx\land\lnot Fx)\lor(\lnot Lx)) \equiv (\lnot Wx \lor Fx \lor \lnot Lx)$, so your formulation is also equivalent to $(1')$. That is, your formulation is correct (in the sense that it has the same logical consequences as the sentence in English), though it was difficult for me to see this by direct translation.

Another equivalent way to translate the first sentence is $$ B \to \lnot(\exists x)(Wx \land Lx \land \lnot Fx). \tag{$1''$}$$

The rest of your formulation is obviously correct: $$(\exists x)(Wx \land Fx) \tag2$$ $$\lnot B \tag3$$

As has been observed in comments, this system is satisfied by a model in which Bluenose is guilty, there is exactly one witness, and the witness is fearful.

I think you were intended to translate the first sentence as follows: $$ B \to \lnot(\exists x)(Wx \land (\lnot Lx \lor Fx)). \tag{1a}$$

This might be the meaning of the first sentence if it had the following logical structure:

If (Bluenose is guilty) then no witness is (lying unless he is fearful).

But that is not the way the English language works. A more logic-friendly phrasing of the English sentence is this:

If (Bluenose is guilty) then (no witness is lying) unless (every witness who lied is fearful).

Take away the parentheses and the sentence still means the same thing.

David K
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  • something did feel off, but I wasn't sure whether it's just some English word play or not. – AlvinL Jul 16 '18 at 13:20
  • The way I arrived at 1 was by considering negation. not (no witness is lying unless fearful) = exists witness such that not (lying unless fearful) and then sort out the unless operation. – AlvinL Jul 16 '18 at 13:25
  • @AlvinLepik I think you interpreted the question the way it was meant to be interpreted. I just think the person who posed the question did a poor job of expressing their intended meaning in English. In English, the "unless" phrase is an alternative to "no witness is lying," not quantified by "no witness". I added my guess as to the intended interpretation (though I believe it does not agree with the meaning in English), and only noticed afterward that it is identical to your formulation. – David K Jul 16 '18 at 13:32